If $X$ and $Y_i$ for $\ 1\le i\le n$ are independent centered Gaussian r.v.'s then do we have $\overset{\rightarrow}X=(X,X+Y_1,X+Y_2,\dots,X+Y_n)$ a Gaussian vector.
Sum of 2 independent Gaussian r.v. is Gaussian so each entry is Gaussian, but there are cases where the vector is not Gaussian itself.
I am given as the definition that $\overset{\rightarrow}X$ is a multivariate Gaussian vector if and only if all linear combinations $∑_{j=1}^d α_jX_j$, $(α_j ∈ R)$are (univariate) Gaussian, but not converse.
on the internet I came across another definition which states,
Let $X := (X_1 ,..., X_n)$ be a random vector. We say that X is a Gaussian random vector if we can write $X = µ + AZ$, where $µ ∈ \mathbb R^n$, $A$ is an $n × k$ matrix and $Z := (Z_1 ,...,Z_k)$ is a k-vector of i.i.d. standard normal random variables.
So is it enough if I take $A=\begin{bmatrix}1&0&0&...\\1&1&0&0&..\\1&0&1&0&0&...\\1&0&0&1&0&...\\...\\1&0&0&0\dots&0&1\end{bmatrix}$ and $Z=(X,Y_1,Y_2,\dots,Y_n)^{T}$
the mean $\mu$ would be again $0$ in this case ?